1. ## finding length

A Steel chimney is held in position by several guy wires. Two such wires AB and AC where B,C and the base of the chimney,D,are on the same level in a staight line with B and C on the left of the chimney. If Angle ABC=45 degrees, ACD=67 degrees and BC=32m. Find the hight of A above D?

2. Originally Posted by zak_killcity14@hotmail.co
A Steel chimney is held in position by several guy wires. Two such wires AB and AC where B,C and the base of the chimney,D,are on the same level in a staight line with B and C on the left of the chimney. If Angle ABC=45 degrees, ACD=67 degrees and BC=32m. Find the hight of A above D?
the figure is like this:
make a right angled triangle, right angled at D
label one vertex as A, other as B
On DB, choose a point C
Now, angleABD(or ABC)=45degrees
and angleACD=67degrees
BC=32 m
from triangle ACD
$\displaystyle \frac{AD}{CD}=tan67$
or$\displaystyle CD=\frac{AD}{tan67}......(1)$
from triangle ABD
$\displaystyle \frac{AD}{BD}=tan45$
or$\displaystyle BD=\frac{AD}{tan45}......(2)$
Subtracting (1) from (2)
$\displaystyle BD-CD=\frac{AD}{tan45} - \frac{AD}{tan67}$
but BD-CD=BC=32
hence
$\displaystyle 32=\frac{AD}{tan45} - \frac{AD}{tan67}$
$\displaystyle AD=$$\displaystyle \frac{32}{\frac{1}{tan45} - \frac{1}{tan67}}$

Keep Smiling
Malay

3. Hello, Zak!

I assume you have a sketch . . .

A steel chimney is held in position by several guy wires.
Two such wires $\displaystyle AB$ and $\displaystyle AC$ where $\displaystyle B,\,C$ and the base of the chimney, $\displaystyle D$,
are on the same level in a straight line with $\displaystyle B$ and $\displaystyle C$ on the left of the chimney.
If $\displaystyle \text{angle }ABC = 45^o,\text{ angle }ACD = 67^o,\;BC = 32\text{ m,}$
find the height of the chimney, $\displaystyle AD.$
Code:
                              * A
* *|
*   * |
*     *  |
*       *   | y
*         *    |
*           *     |
* 45°         * 67°  |
*---------------*-------*
B      32       C   x   D

Let $\displaystyle x = CD,\:y = AD.$

In right triangle $\displaystyle ADB:\;\tan45^o = \frac{y}{x+32}\quad\Rightarrow\quad1 = \frac{y}{x+32}\quad\Rightarrow\quad x \:= \:y - 32$ [1]

In right triangle $\displaystyle ADC:\;\tan67^o = \frac{y}{x}\quad\Rightarrow\quad x \,=\,\frac{y}{\tan67^o}$ [2]

Equate [1] and [2]: .$\displaystyle y - 32 \:=\:\frac{y}{\tan67^o}$

Solve for $\displaystyle y:\;\;y\tan67^o - 32\tan67^o \:=\:y\quad\Rightarrow\quad y\tan67^o - 32\tan67^o \:=\:y$

. . $\displaystyle y\tan67^o - y \:=\:32\tan67^o\quad \Rightarrow\quad y(\tan67^o - 1) \:=\:32\tan67^o$

Therefore: .$\displaystyle y \;=\;\frac{32\tan67^o}{\tan67^o - 1} \;=\;55.60138965 \;\approx\;\boxed{55.6\text{ m}}$